Problem: Let $h(x)=x^3e^{-x}$. What is the absolute maximum value of $h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{e^3}$ (Choice B) B $\dfrac{e^3}{3}$ (Choice C) C $\dfrac{27}{e^3}$ (Choice D) D $h$ has no maximum value
Answer: Let's first find the relative extremum points of $h$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=x^2e^{-x}(3-x)$. $h'(x)=0$ for $x=0,3$. $h'$ is defined for all real numbers. Therefore, our only critical points are $x=0,3$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x<0$ $0<x<3$ $x>3$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<0$ $x=-1$ $h'(-1)=4e>0$ $h$ is increasing $\nearrow$ $0<x<3$ $x=1$ $h'\left(1\right)=\dfrac2e>0$ $h$ is increasing $\nearrow$ $x>3$ $x=4$ $h'(4)=-\dfrac{16}{e^4}<0$ $h$ is decreasing $\searrow$ Let's imagine ourselves walking on the graph of $h$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up and up until we reach $x=3$. Then, we will be forever going down. Therefore, $h$ must obtain its absolute maximum value at $x=3$. We are asked to find that maximum value, which is $h\left(3\right)=\dfrac{27}{e^3}$. In conclusion, the absolute maximum value of $h$ is $\dfrac{27}{e^3}$.